Trivial contradiction to the 3rd postulate

1. Jun 22, 2004

Suppose we have two operators A and B, such that they are both hermitian with eigenvalues ai and bi. We construct a new operator like this C=A +iB, where i is the imaginary unity. Operator C is observable in the sense that it is measurable. Yes, if we measure A and B (suppose for simplicity that they commute) we get that the measured value of C is a+ib. Thus we obtain an OBSERVABLE with complex eigenvalues, in contradiction with the 3rd postulates wich states that all observable quantities are associated with hermitian operators with complete bases of eigenstates, bla bla. Kinda stupid but it works.

2. Jun 22, 2004

zefram_c

This is already a problem, as hermitian operators must have real eigenvalues. Still, that condition can be relaxed.

No, this operator is NOT observable. It cannot correspond to a physical observable since you *defined* it to be complex, assuming A and B have real, nonzero eigenvalues. It is an operator with well defined complex eigenvalue solutions, but it cannot be directly measured. Measuring A and B and determining C that way is not the same as measuring C. Furthermore, it's easy to see that C is not hermitian since the hermitian conjugate of A+iB is A-iB (assuming that A and B are both hermitian)

3. Jun 22, 2004

Eye_in_the_Sky

Hello, tavi_boada. Tell me if this helps.

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Consider a nondegenerate hermitian operator A which corresponds to an observable of some dynamical attribute of the quantum system. Let A have eigenvectors |n> and corresponding eigenvalues a(n), where n = 1,2, ... . Write A as

A = Sigma_n { a(n)|n><n| } .

The numbers a(n) correspond to possible measured values of a dynamical attribute of the system. As such, they are necessarily real numbers. On the other hand, nothing stops me from going to the "pointer" of the measuring device and changing the numbers which the "pointer" points at. Let's say I take a(n) --> b(n) in a one-to-one fashion. I then have a new operator

B = Sigma_n { b(n)|n><n| } .

The newly obtained measuring device is a "real physical device", and, in that sense, the associated operator B corresponds to something "measurable". However, the numbers b(n) may not represent possible values of a dynamical attribute of the system, as did the original a(n). In particular, the b(n) can even be complex numbers.

Now, since the values of the dynamical attributes of physical systems are necessarily real numbers, and the physical quantities which we measure are necessarily real-valued functions of those numbers, the term "observable" has been reserved for an operator which necessarily has real eigenvalues. This means that, while our operator B above, with complex b(n), does in some sense correspond to something "measurable", it is not an "observable".

Last edited: Jun 22, 2004
4. Jun 22, 2004

shchr

Hello, tavi_boada. I'm interested in your assertion.
But I think an observable corresponding to a physical value must be a one which can be observed by direct mesurement. But, although A and B are directly observed, A+iB cannot be observed directly. Could you give us an example of an observable (directly observalbe value) like A+iB.
Thank you.

Last edited: Jun 22, 2004
5. Jun 22, 2004

Eye_in_the_Sky

This time I'll answer your question directly.

Yes, your operator C is an "observable" in the sense that you give. Its eigenvalues are like a "code" for those of A and B.

At the time when physicists defined the notion of an "observable" they weren't thinking in the most general terms that you have suggested. They weren't thinking of an eigenvalue as an "abstract label", but rather, as the "numerical value of a physical quantity". So, they went ahead defined an "observable" as having real eigenvalues.

6. Jun 23, 2004

Hey guys thank you for your interesting contributions.

The question is, are observables things you can measure with a device, like pressure, position?Or are they al, in general,l REAL functions of these?The hamiltonian operator has eigenvalues that are the energies the system can take. Energy is not something you can directly measure. There are no energyometers. I view the hamiltonian as an operator real function of other operators.

Ther third postulate as you know says that all observable quantities "correspond to" a hermitian operator.THe hamiltonian operator is a real function of hermitian operators so it is also hermitian. But complex functions of hermitian operators are not in general hermitian.

If we consider that a real function of observables is observable, then a complex function of them is also observable. No?

7. Jun 23, 2004

shchr

Let's take the case of harmonic oscillator. x is an observable and p is also an observable. But they cannot be measured simultaneously because they don't commute. As we know, a=C x + i D p (C and D are proper constants) is annihilation operator. But a does not have diagonal elements which should be average values of some physical quantity. So a must not be an observable. As far as I know, there is not such a combination of observables which are also obsevables but are not real values.

8. Jun 23, 2004

slyboy

It is better to use the projector formalism for quantum measurements in order to get around tricky cases like these. Observables fall into equivalence classes, that have the same projectors onto their eigenspaces. The projectors are treated as fundamental, rather than the observables.